Find ways from $(0,0)$ to $(8,8)$ You are allowed only to go east or north. Because of road construction, you cannot touch the points $a, b, c$ and $d$. Under these restrictions, the number of ways that you can go from $(0, 0)$ and finish at $(8, 8)$ in the following figure is:

First, I used ${16\choose 8}$ to get the number of ways without this construction. I am a bit confused on what to do next...
 A: Through $a$. $\binom{6}{3}\times\binom{10}{5}$.
Through $b$ without going through $a$. $\binom{6}{4}\times\binom{9}{4}$.
Through $d$ without going through $a$. $\binom{6}{2}\times\binom{9}{5}$.
$\binom{16}{8}-\binom{6}{3}\times\binom{10}{5}-\binom{6}{4}\times\binom{9}{4}-\binom{6}{2}\times\binom{9}{5}=4050$
A: Here's how to solve this:
You are correct that there are ${16 \choose 8}$ paths without restrictions.
Now, consider paths that go through $a$.  How many distinct initial path segments get you to $a$?  There are ${6 \choose 3}$ of them, using the same logic of your broader solution.  For each of these initial path segments, how many ways could you pass (ignoring the constraint)?  There are ${10 \choose 5}$ of them (again by your logic).  How many of these final paths go through $b$?  Half of them!  How many of these final paths go through $d$?  The other half of them!  These are all precluded.  So now you know how many of the total number of paths are prevented, because they go through $a$.
Now, consider $d$.  How many paths go through $d$ *but did not go through $a$?  For that to happen, the path would have had to go through $(2,4)$, and then head east.  (Do you see why?).  Thus, use the same method to compute the number of paths that go through $(2,4)$ and then go east.  For each of these, you can determine the number of unrestricted paths from $d$.
Do you get the basic logic here?  Can you continue on your own?
